JEE Advanced · Mathematics · 32. Probability
Let \(E^c\) denotes the complement of an event \(E\). Let \(E, F, G\) be pairwise independent events with \(P(G)>0\) and \(P(E \cap F \cap G)=0\). Then, \(P\left(E^c \cap F^c \mid G\right)\) equals
- A
\(P\left(E^c\right)+P\left(F^c\right)\)
- B
\(P\left(E^c\right)-P\left(F^c\right)\)
- C
\(P\left(E^c\right)-P(F)\)
- D
\(P(E)-P\left(F^c\right)\)
Answer & Solution
Correct Answer
(C)
\(P\left(E^c\right)-P(F)\)
Step-by-step Solution
Detailed explanation
\[
\text { } \begin{aligned}
P\left(\frac{E^c \cap F^c}{G}\right) & =\frac{P\left(E^c \cap F^c \cap G\right)}{P(G)} \\
& =\frac{P(G)-P(E \cap G)-P(G \cap F)}{P(G)} \\
& =\frac{P(G)[1-P(E)-P(F)]}{P(G)} \\
& =1-P(E)-P(F) \\
& =P\left(E^c\right)-P(F) .
\end{aligned}
\]
\text { } \begin{aligned}
P\left(\frac{E^c \cap F^c}{G}\right) & =\frac{P\left(E^c \cap F^c \cap G\right)}{P(G)} \\
& =\frac{P(G)-P(E \cap G)-P(G \cap F)}{P(G)} \\
& =\frac{P(G)[1-P(E)-P(F)]}{P(G)} \\
& =1-P(E)-P(F) \\
& =P\left(E^c\right)-P(F) .
\end{aligned}
\]
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